A thin circular ring of mass $M$ and radius '$r$' is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be: (2003)

$\frac{M\omega}{4m}$

$\frac{M\omega}{M + 4m}$

$\frac{(M + 4m)\omega}{M}$

$\frac{(M + 4m)\omega}{M + 4m}$

Solution:

By conservation of angular momentum, $I_{initial}\omega_{initial} = I_{final}\omega_{final}$.
Initial moment of inertia $I_i = Mr^2$. Final moment of inertia $I_f = Mr^2 + 4mr^2$.
Thus, $$Mr^2 \omega = (M + 4m)r^2 \omega' \implies \omega' = \frac{M\omega}{M + 4m}$$.

Rankers Physics

A thin circular ring of mass $M$ and radius '$r$' is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be: (2003)

Solution: